| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Moderate -0.3 This is a straightforward application of the impulse-momentum theorem with vectors. Students must find the change in momentum (mass × change in velocity), then calculate the magnitude using Pythagoras. It requires only direct formula application with no problem-solving insight, making it slightly easier than average for M2. |
| Spec | 6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\mathbf{I} = \Delta\text{mom.} = 0.5[(13\mathbf{i} + 7\mathbf{j}) - (5\mathbf{i} - 8\mathbf{j})]\) | M1 A1 | |
| \(= 0.5(8\mathbf{i} + 15\mathbf{j})\) | A1 | |
| mag. of \(\mathbf{I} = 0.5\sqrt{8^2 + 15^2} = 8.5\) Ns | M1 A1 | (5) |
## Question 1:
| Answer/Working | Marks | Notes |
|---|---|---|
| $\mathbf{I} = \Delta\text{mom.} = 0.5[(13\mathbf{i} + 7\mathbf{j}) - (5\mathbf{i} - 8\mathbf{j})]$ | M1 A1 | |
| $= 0.5(8\mathbf{i} + 15\mathbf{j})$ | A1 | |
| mag. of $\mathbf{I} = 0.5\sqrt{8^2 + 15^2} = 8.5$ Ns | M1 A1 | **(5)** |
---\begin{enumerate}
\item An ice hockey puck of mass 0.5 kg is moving with velocity $( 5 \mathbf { i } - 8 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular horizontal unit vectors, when it is struck by a stick. After the impact, the puck travels with velocity $( 13 \mathbf { i } + 7 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\end{enumerate}
Find the magnitude of the impulse exerted by the stick on the puck.\\
(5 marks)\\
\hfill \mbox{\textit{Edexcel M2 Q1 [5]}}